I just started to study tensor products and have a question. Suppose A is a (commutative with unity) ring, B an A-algebra and M an A-module. Consider map $\alpha: M \rightarrow B \otimes_AM$, defined by $m \mapsto 1 \otimes m$. My question is, when is $\alpha$ injective?

I thought about this particular case: $\alpha: \mathbb{Z} \rightarrow \mathbb{F}_2 \otimes_\mathbb{Z} \mathbb{Z}$. It seems to me that $\alpha(2z) = 0$, for every $z \in \mathbb{Z}$, so $\alpha$ is not injective in this case.

But can we prove that if $B$ is a ring containing $A$ as a subring, then $\alpha$ is injective?

  • $\begingroup$ Note that $B\otimes_A A\cong B$ under map $b\otimes a\mapsto ba$ in general (i.e. $A$ acts as neutral element under tensoring with respect to $A$). But, you've done well to notice that trouble lies in characteristic. I'm not sure how much can be said about injectivity in general. $\endgroup$ – Ennar Aug 24 '16 at 13:49
  • $\begingroup$ This is a dupe of math.stackexchange.com/questions/505345/… so we should close one of them when the other is answered $\endgroup$ – Matthew Towers Aug 24 '16 at 14:00
  • $\begingroup$ Thank you! How about this: $\mathbb{F}_2 \rightarrow \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{F}_2$? Every element of $\mathbb{F}_2$ has finite (additive) order, so it must be sent to $0$, since any element of a vector space over $\mathbb{Q}$ is of infinite order. Is that right? $\endgroup$ – l_j Aug 24 '16 at 14:21

When $A$ is a subring of $B$, you question is equivalent to:

Given the exact sequence $0\rightarrow A\rightarrow B$, is $0\rightarrow A\otimes_AM=M\rightarrow B\otimes_AM$ exact.

There is a notion of flatness in commutative algebra, an $A$-module $M$ is flat if and only for every injective map $ N\rightarrow N'$, $N\otimes_AM\rightarrow N'\otimes_AM$ is injective.

They are many conditions which insure that a module is flat for example if $M$ is finitely generated and $p$ is a prime ideal of $A$, the localization $M_p$ is a free $A_p$, module. A partial answer to your question will be if $M_p$ is a free $A_p$ module for every prime ideal $p$ of $A$, $M\rightarrow B\otimes_AM$ is injective.


  • $\begingroup$ Thank you! I'll look into flat modules. $\endgroup$ – l_j Aug 24 '16 at 15:49

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